Question 651018

to  determine whether , plug in values for {{{x=1}}}, {{{y=-5}}} and {{{z=10}}}. I you get that left side of equation is equal to right side, than (1,-5,10) is a solution of the following system

{{{x+y+z=6}}}..->..{{{1-5+10=6}}}->..{{{11-5=6}}}->..{{{6=6}}}

{{{3x-y-z=-2}}}..->..{{{3*1-(-5)-10=-2}}}->..{{{3+5-10=-2}}}->..

{{{8-10=-2}}}->..{{{-2=-2}}}

{{{2x-y+4z=37}}}..->..{{{2*1-(-5)+4*10=37}}}.->..{{{2+5+40=37}}}->..{{{47=37}}}...not true

so, the ordered triple ({{{1}}}, {{{-5}}}, {{{10}}}) is NOT a solution of this system


here is a solution:


*[invoke cramers_rule_3x3 1, 1, 1, 6, 3, -1, -1, -2, 2, -1, 4, 37]


as you can see, the three solutions are {{{x=1}}}, {{{y=-3}}}, and {{{z=8}}} giving the ordered triple ({{{1}}}, {{{-3}}}, {{{8}}})